\section{Suggested solutions to unequal error protection}

The video stream used in this project is MPEG-4 encoded, which means that the video is divided into interframes (I-frames) and intraframes (P-frames). The parts of the video crucual to the multimedia experience is the I-frames because

\subsection{Solution 1}

The data that needs protection is identified in \fixme{B: Insert ref} to be the I-frames of the video stream. To protect the I-frames the source packet matrix, $\overline{\overline{M}}$, consists of $a$ NC'ed I-frame packets and $g-a$ NC'ed packets of I-frame redundancy and P-frame combined. 


\begin{align} &
\overline{\overline{M}}
=
\left[
\begin{array}{cc}	
\begin{bmatrix}
i_{0,0}	&	\hdots 	& i_{0,b} \\
\vdots	&	\ddots	& \vdots\\
i_{a,0}	&	\hdots 	& i_{a,b} \\
\end{bmatrix}	 & 0 \\
\begin{bmatrix}
r_{a+1,0}	&	\hdots 	& r_{a+1,b} \\
\vdots	&	\ddots	& \vdots\\
r_{g,0}	&	\hdots 	& r_{g,b} \\
\end{bmatrix}	 &  
\begin{bmatrix}
p_{a+1,b+1}	&	\hdots 	& p_{a+1,m} \\
\vdots	&	\ddots	& \vdots\\
p_{g,0}	&	\hdots 	& p_{g,m} \\
\end{bmatrix}
\end{array}
\right]
\label{eq:solution1-source-matrix}
\intertext{Where:}
&\text{$\overline{\overline{M}}$ is the $g\times m$ source packet matrix} \notag\\
&\text{$g$ is the generation size} \notag\\
&\text{$m$ is the symbol length} \notag\\
&\text{$i_{a,b}$ is I-frame byte with index $(a,b)$} \notag\\
&\text{$r_{g,b}$ is the I-frame redundancy byte with index $(g,b)$} \notag\\
&\text{$p_{g,m}$ is the P-frame byte with index $(g,m)$} \notag\\
\end{align}


The idea is to send the exact number of needed I-frame packets as the first part of the generation and then combining P-frame and I-frame redundancy in the last part. In this way the packets will very often bear new information and there is a possibility to correct errors in the I-frame packets from the redundancy even though the P-frame part isn't solvable. 
In that way the I-frame and P-frame is unequally error protected because that probability of solving the I-frame should be highest. 

Sandsynligheden for at I-framen kan løses når både I- og P-delen er blevet sendt skal findes. For at finde den skal man finde sandsynligheden for at alle I-delene er blevet modtaget samt sandsynligheden for at I-framen kan dekodes ud fra I og redundancy. Desuden skal sandsynligheden for at P-delen kan elimineres så man rent faktisk kan bruge redundancy til noget beregnes.
Det er jeg ikke i stand til.